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Svetllana [295]
3 years ago
11

Marcar la alternativa correcta en relación a desplazamiento y la distancia

Chemistry
1 answer:
asambeis [7]3 years ago
8 0

Answer              A

Explanation:

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Select the correct answer from each drop-down menu.
stealth61 [152]

Answer:

Light energy; chemical

Explanation:

The light energy of the Sun is converted into chemical energy by the plants.

5 0
3 years ago
How many neutrons are in selenium -36?
Soloha48 [4]

Selenium (Se)  the most common isotope of this element. The nucleus consists of 34 protons (red) and 46 neutrons (blue).

5 0
3 years ago
How many mL of a 1.48 M calcium hydroxide solution are needed to neutralize 36.0 mL of a 1.63 M hydrochloric acid solution
Lelechka [254]

The volume (in mL) of calcium hydroxide, Ca(OH)₂ needed for the reaction is 19.8 mL

<h3>Balanced equation </h3>

2HCl + Ca(OH)₂ —> CaCl₂ + 2H₂O

From the balanced equation above,

  • The mole ratio of the acid, HCl (nA) = 2
  • The mole ratio of the base, Ca(OH)₂ (nB) = 1

<h3>How to determine the volume of Ca(OH)₂ </h3>
  • Molarity of base, Ca(OH)₂ (Mb) = 1.48 M
  • Volume of acid, HCl (Va) = 36 mL
  • Molarity of acid, HCl (Ma) = 1.63 M
  • Volume of base, Ca(OH)₂ (Vb) =?

MaVa / MbVb = nA / nB

(1.63 × 36) / (1.48 × Vb) = 2

58.68 / (1.48 × Vb) = 2

Cross multiply

2 × 1.48 × Vb = 58.68

2.96 × Vb = 58.68

Divide both side by 2.96

Vb = 58.68 / 2.96

Vb = 19.8 mL

Learn more about titration:

brainly.com/question/14356286

#SPJ1

6 0
2 years ago
Gold has a density of 1,200 lb./ft. What is the density of gold in g/em? For conversion factors use I lb. 453.6 g, and l inch-2.
Jlenok [28]

<u>Answer:</u> The density of gold in g/cm^3 is 19.22g/cm^3

<u>Explanation:</u>

Density is defined as the ratio of mass of the object and volume of the object. Mathematically,

\text{Density}=\frac{\text{Mass of the object}}{\text{Volume of the object}}

We are given:

Density of gold = 1200lb/ft^3

Using conversion factors:

1 lb = 453.6 g

1 feet = 12 inches

1 inch = 2.54 cm

Converting given quantity into g/cm^3, we get:

\Rightarrow (\frac{1200lb}{ft^3})\times (\frac{453.6g}{1lb})\times (\frac{1ft}{12inch})^3\times (\frac{1inch}{2.54cm})^3\\\\\Rightarrow 19.22g/cm^3

Hence, the density of gold in g/cm^3 is 19.22g/cm^3

6 0
3 years ago
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